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2 years ago

Assuming that both cases describe hydrogen‑like atoms with one electron, for which case is more energy emitted or absorbed?

1 answer:

5 0

Assuming that both cases describe hydrogen‑like atoms with one electron, More energy is emitted or absorbed for case 2. The correct option is D.

<h3>What is emitting of energy, by electron?</h3>

The energy of the electron decreases as it changes levels, and emission of photons happens in the atom.

With the electron moving from a higher to a lower energy level, the photon is emitted. The photon's energy is the same as the energy lost by an electron moving to a lower energy level.

Thus, the correct option is D, More energy is emitted or absorbed for case 2.

Learn more about emitting of energy

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Hello, please help me if you can :(

KIM [24]

Answer:

4ml

Explanation:

6 0

4 years ago

15 points + brainliest pls help :(

Komok [63]

Answer:

27 g

Explanation:

M(C6H12O6) = 6*12 + 12*1 + 6*16 = 180 g/mol

100 mL = 0.1 L solution

1.5 M = 1.5 mol/L

1.5 mol/L * 0.1 L = 0.15 mol C6H12O6

0.15 mol * 180 g/1 mol = 27 g C6H12O6

6 0

3 years ago

In the laboratory a general chemistry student finds that when 2.84 g of KClO4(s) are dissolved in 107.70 g of water, the tempera

vredina [299]

Answer : The enthalpy change of dissolution of $KClO_4$ is 54.3 kJ/mole

Explanation :

Heat released by the reaction = Heat absorbed by the calorimeter + Heat absorbed by the water

$q=[q_1+q_2]$

$q=[c_1\times \Delta T+m_2\times c_2\times \Delta T]$

where,

q = heat released by the reaction

$q_1$ = heat absorbed by the calorimeter

$q_2$ = heat absorbed by the water

$c_1$ = specific heat of calorimeter = $1.55J/^oC$

$c_2$ = specific heat of water = $4.184J/g^oC$

$m_2$ = mass of water = 107.70 g

$\Delta T$ = change in temperature = $T_2-T_1=(22.80-20.34)=2.46^oC$

Now put all the given values in the above formula, we get:

$q=[(1.55J/^oC\times 2.46^oC)+(107.70g\times 4.184J/g^oC\times 2.46^oC)]$

$q=1112.3J=1.1123kJ$

Now we have to calculate the enthalpy change of dissolution of $KClO_4$

$\Delta H=\frac{q}{n}$

where,

$\Delta H$ = enthalpy change = ?

q = heat released = 1.1123 kJ

m = mass of $KClO_4$ = 2.84 g

Molar mass of $KClO_4$ = 138.55 g/mol

$\text{Moles of }KClO_4=\frac{\text{Mass of }KClO_4}{\text{Molar mass of }KClO_4}=\frac{2.84g}{138.55g/mole}=0.0205mole$

$\Delta H=\frac{1.1123kJ}{0.0205mole}=54.3kJ/mole$

Therefore, the enthalpy change of dissolution of $KClO_4$ is 54.3 kJ/mole

5 0

4 years ago

Friction. That was the topic Sally and Sue are investigating in science. They used a small car and a ramp as seen in the

DaniilM [7]

Answer: A: The smoother the ramp surface, the greater the car's speed.

Explanation: I did the USA test prep

3 0

3 years ago

g The difference between an ionic bond and a covalent bond is that Select one: a. ionic bonds commonly occur between two metals

Bond [772]

Answer:

C) Covalent bonds come about because of a sharing of electrons; ionic bonds do not.

Explanation:

There are two main types of chemical bonds- covalent and ionic/electrovalent bonds.

Ionic bond: Ionic or electrovalent bonds are characterized by the transfer of electrons from electropositive atoms (metals) to electronegative atoms (non-metals). The metal atoms after donating their electrons become positively charged ions (cations) while the non-metal atoms after accepting electrons become negatively charged ions (anions). Strong electrostatic forces of attraction constitutes ionic bonds.

Covalent bond: Covalent bonds are formed by the sharing of electrons by the atoms involved in the bond; usually between atoms of comparable electronegativities or atoms of the same element. The shared electrons are contributed by each of the atoms involved in the bonding or may be contributed by only one of the atoms. In covalent bonding, molecules rather than ions are formed.

8 0

3 years ago